How do you find a linear approximation to #root(4)(84)# ?

Note that #3^4 = 81#, which is close to #84#.

So #root(4)(84)# is a little larger than #3#.

To get a better approximation, we can use a linear approximation, a.k.a. Newton's method.

Define:

#f(x) = x^4-84#

Then:

#f'(x) = 4x^3#

and given an approximate zero #x=a# of #f(x)#, a better approximation is:

#a - (f(a))/(f'(a))#

So in our case, putting #a=3#, a better approximation is:

#3-(f(3))/(f'(3)) = 3-(3^4-84)/(4(3)^3) = 3-(81-84)/(4 * 27) = 3+1/36 = 109/36 = 3.02bar(7)#

This is almost accurate to #4# significant figures, but let's quote the approximation as #3.03#

Note that the linear approximation near a point #a# can be given by:

#f(x) ~~f(a)+f'(a)(x-a)#

If given: #f(x) = root(4)(x)#

then a suitable choice for #a# would be #a=81# because we know #root(4)81=3# exactly and it is close to #84#.

So:

#f(a) = f(81) = root(4)(81)=3 #

Also;

#f(x) = x^(1/4)# so #f'(x) = 1/4x^(-3/4)=1/(4root(4)(x)^3)#

#f'(81) = 1/(4root(4)(81)^3)=1/(4*3^3)=1/108#

Therefore we can approximate (near #81#):

#f(x)~~f(a)+f'(a)(x-a)#

#implies root(4)(x)~~3+1/(108)(x-81)#

So:

#root(4)(84)=3+1/108(84-81)#

#3+1/108*3=324/3+3/108=327/108~~3.02778#

The more accurate value is #3.02740#

so the linear approximation is fairly close.

We can say that we have a function of #f(x)= root(4) (x)#
and # root(4) (84)=f(84)#

Now, let's find the derivative of our function.

We use the power rule, which states that if #f(x)=x^n#, then #f'(x)=nx^(n-1)# where #n# is a constant.

#f(x)=x^(1/4)#

=>#f'(x)=1/4*x^(1/4-1)#

=>#f'(x)=(x^(-3/4))/4#

=>#f'(x)=1/x^(3/4)*1/4#

=>#f'(x)=1/(4x^(3/4))#

Now, to approximate # root(4) (84)#, we try to find the perfect fourth-power closest to 84

Let's see...
#1#
#16#
#81#
#256#
.
.
.
We see that #81# is our closest one.

We now find the tangent line of our function when #x=81#

=>#f'(81)=1/(4*81^(3/4))#

=>#f'(81)=1/(4*81^(2/4)*81^(1/4))#

=>#f'(81)=1/(4*9*3)#

=>#f'(81)=1/108#

This is the slope we are looking for.

Let's try to write the equation of the tangent line in the form #y=mx+b#

Well, what is #y# equal to when #x=81#?

Let's see...
#f(81)=root(4)(81)#
=>#f(81)=3#

Therefore, we now have:
#3=m81+b# We know that the slope, #m#, is #1/108#

=>#3=1/108*81+b# We can now solve for #b#.

=>#3=81/108+b#

=>#3=3/4+b#

=>#2 1/4=b#

Therefore, the equation of the tangent line is #y=1/108x+2 1/4#

We now use 84 in the place of #x#.

=>#y=1/108*84+2 1/4#

=>#y=1/9*7+2 1/4#

=>#y=7/9+9/4#

=>#y=28/36+81/36#

=>#y=109/36#

=>#y=3.02bar7#

Therefore, #root [4](84)~~3.02bar7#